Ordering on Natural Numbers is Compatible with Addition/Corollary
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Corollary to Ordering on Natural Numbers is Compatible with Addition
Let $a, b, c, d \in \N$ where $\N$ is the set of natural numbers.
Then:
- $a > b, c > d \implies a + c > b + d$
Proof
| \(\displaystyle a\) | \(>\) | \(\displaystyle b\) | |||||||||||
| \((1):\quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle a + c\) | \(>\) | \(\displaystyle b + c\) | Ordering on Natural Numbers is Compatible with Addition |
| \(\displaystyle c\) | \(>\) | \(\displaystyle d\) | |||||||||||
| \((2):\quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle b + c\) | \(>\) | \(\displaystyle b + d\) | Ordering on Natural Numbers is Compatible with Addition |
Finally:
| \(\displaystyle a + c\) | \(>\) | \(\displaystyle b + c\) | from $(1)$ | ||||||||||
| \(\displaystyle b + c\) | \(>\) | \(\displaystyle b + d\) | from $(2)$ | ||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle a + c\) | \(>\) | \(\displaystyle b + d\) | Ordering on Natural Numbers is Trichotomy |
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: I. Basic Concepts ... (previous) ... (next): Introduction $\S 4$: The natural numbers: Exercise $1$
